Yesterday we published Ivy Bridge's transistor count as 1.48 billion. It turns out that was wrong as Intel's Mooly Eden accidentally read the B in billion as an 8 while on stage. The real number is 1.4 billion. When we published that story we compared it to Intel's Sandy Bridge, which at launch was said to be 995 million transistors. However at IDF Intel had been using another number: 1.16 billion transistors. It turns out both are right, but why is there a difference?

When designing a microprocessor you end up with a schematic of all of the circuits and transistors in the design. With the design schematic done layout is next on the list. However sometimes in the process of moving from the schematic to layout phase, transistor count baloons. The reason is simple. There are some circuits which may be represented by a single transistor at the schematic phase, but for more efficient layout use four transistors in tandem. For Sandy Bridge the 995M number is for the number of transistors in the schematic, while 1.16 billion is how many transistors are put down at the fab. Both are correct, but the 1.16B number is directly comparable to Ivy Bridge's 1.4B transistors.

That puts Ivy Bridge's transistor count at 20.7% higher than Sandy Bridge, which is more in line with what to expect from a tick.

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  • ssj4Gogeta - Thursday, September 15, 2011 - link

    He's not "expecting" or predicting anything. He's just stating what Intel _told_ him. Reply
  • jwilliams4200 - Thursday, September 15, 2011 - link

    Did you not read the article?

    "That puts Ivy Bridge's transistor count at 20.7% higher than Sandy Bridge, which is more in line with what to EXPECT from a tick."

    It does NOT say that Intel told him to expect a 20% increase.
    Reply
  • ssj4Gogeta - Thursday, September 15, 2011 - link

    "It turns out that was wrong as Intel's Mooly Eden accidentally read the B in billion as an 8 while on stage. The real number is 1.4 billion."

    1.4/1.16 = 1.206
    Reply
  • jwilliams4200 - Thursday, September 15, 2011 - link

    Please try to read and UNDERSTAND the article before posting nonsense comments.

    The article was about the difference between what Anand EXPECTED (about 20% increase) and what the actual increase was. Anand was initially surprised about the increase (when he had the wrong counts), but with the corrected counts, he says it is about what he would expect.
    Reply
  • ssj4Gogeta - Thursday, September 15, 2011 - link

    First off, Moore's law doesn't have anything to do with transistor counts. It has to do with transistor density. Why will transistor count magically increase if you're shrinking the same design to a smaller node?

    In fact, the only increase will come from the new features being added. You cannot predict it with math as you're trying to do in your first comment. So you shouldn't really be "expecting" anything particular from the tick unless you know what features are being added. You can either save costs by building the same features on a smaller die or add more features on the same size die, or any combination of the two.

    Conroe (65nm Core 2 Duo) had 291 million transistors on a 143mm² die, while Wolfdale (45nm Core 2 Duo) had 410 million transistors on a 107mm² die. In this case the transistor count increase was 40% (mostly for the bigger L2) and die size was decreased as well, which means Intel opted to go for a combination of saving costs and adding more transistors.

    If you calculate average transistor densities, they're 2.03 million per mm² for Conroe and 3.8 million per mm² for Wolfdale, a near doubling according to Moore's law.
    Reply
  • jwilliams4200 - Thursday, September 15, 2011 - link

    No, you are completely wrong. Moore's Law does refer to transistor counts on ICs. A common statement of the law is:

    "The number of transistors that can be placed inexpensively on an integrated circuit doubles approximately every two years."

    Which still does not answer the question, on what basis does Anand "expect" transistor counts to increase by about 20% for a tick?
    Reply
  • Stahn Aileron - Thursday, September 15, 2011 - link

    BTW, the quote you supplied in fact has to deal with DENSITY, not actual count.

    Moore's Law is about how much you can pack within a given area. It doesn't NOT necessarily require an increase in transistor count for a given IC design.

    You're looking at different sides of the same coin. Neither one of of you are wrong.
    Reply
  • Stahn Aileron - Thursday, September 15, 2011 - link

    *It does NOT necessarily... Reply
  • jwilliams4200 - Thursday, September 15, 2011 - link

    Density is number of transistors per square millimeter (or some other unit). That is NOT what the statement of Moore's law that I quoted says. Reply
  • ssj4Gogeta - Thursday, September 15, 2011 - link

    In fact, that is what it means.

    "The number of transistors that _CAN_ be placed _inexpensively_ on an integrated circuit doubles approximately every two years."

    Notice the word "can". It doesn't mean manufacturers have to keep doubling the transistor counts. It just means that they can do so for the same cost. In other words, transistors become half as costly with every step. That is, the transistor density keeps doubling and so the amount of silicon needed keeps getting halved.
    Reply

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